Not only AM-GM 4 (The Other AM-GM Part 6)

Algebra Level 5

Positive real numbers a a , b b , c c , and d d are such that a = 27 c a = 27 - c and 1 b = 21 1 d \dfrac{1}{b} = 21 - \dfrac{1}{d} . Find the maximum value of the expression below to 3 decimal places.

1 1 a + b + 1 c + d \large\dfrac{1}{ \dfrac{1}{a} + b + \dfrac{1}{c} + d}

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The answer is 2.953.

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1 solution

Anirudh Sreekumar
Apr 24, 2017

a + c = 27 ( Given ) 1 b + 1 d = 21 ( Given ) Applying AM HM, We have, a + c 2 2 1 a + 1 c 1 a + 1 c 4 a + c 4 27 Similarly, 1 b + 1 d 2 2 b + d b + d 4 1 b + 1 d 4 21 1 a + b + 1 c + d 4 27 + 4 21 1 1 a + b + 1 c + d 27 × 21 4 × 48 2.953125 \begin{aligned} a+c&=27 \hspace{7mm}\small\color{#3D99F6} (\text{Given})\\ \dfrac1 b+\dfrac 1 d&=21\hspace{7mm}\small\color{#3D99F6} (\text{Given}) \\ \text{Applying AM}&\geq\text{HM, We have,}\\\\ \dfrac{a+c}{2}&\geq\dfrac{2}{\dfrac1 a+\dfrac 1 c}\\ \implies \dfrac1 a+\dfrac 1 c&\geq\dfrac{4}{a+c}\\ &\geq\dfrac4 {27}\\ \text{Similarly,}\\\\ \dfrac{\dfrac1 b+\dfrac 1 d}{2}&\geq\dfrac{2}{b+d}\\ \implies b+d&\geq\dfrac{4}{\dfrac1 b+\dfrac 1 d}\\ &\geq\dfrac4 {21}\\\\ \implies \dfrac1 a +b+\dfrac1 c +d&\geq\dfrac4 {27}+\dfrac4 {21}\\\\ \implies\dfrac{1}{ \dfrac1 a +b+\dfrac1 c +d}&\leq\dfrac{27\times 21}{4\times48}\\ &\leq2.953125\end{aligned}

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